Đáp án:
$\begin{array}{l}
a)Dk{\rm{xd}}:a \ge 0;a \ne 1\\
A = \dfrac{{3a + \sqrt {9a} - 3}}{{a + \sqrt a - 2}} - \dfrac{{\sqrt a + 1}}{{\sqrt a + 2}} + \dfrac{{\sqrt a - 2}}{{1 - \sqrt a }}\\
= \dfrac{{3a + 3\sqrt a - 3 - \left( {\sqrt a + 1} \right)\left( {\sqrt a - 1} \right) - \left( {\sqrt a - 2} \right)\left( {\sqrt a + 2} \right)}}{{\left( {\sqrt a + 2} \right)\left( {\sqrt a - 1} \right)}}\\
= \dfrac{{3a + 3\sqrt a - 3 - \left( {a - 1} \right) - \left( {a - 4} \right)}}{{\left( {\sqrt a + 2} \right)\left( {\sqrt a - 1} \right)}}\\
= \dfrac{{3a + 3\sqrt a - 3 - a + 1 - a + 4}}{{\left( {\sqrt a + 2} \right)\left( {\sqrt a - 1} \right)}}\\
= \dfrac{{a + 3\sqrt a + 2}}{{\left( {\sqrt a + 2} \right)\left( {\sqrt a - 1} \right)}}\\
= \dfrac{{\left( {\sqrt a + 1} \right)\left( {\sqrt a + 2} \right)}}{{\left( {\sqrt a + 2} \right)\left( {\sqrt a - 1} \right)}}\\
= \dfrac{{\sqrt a + 1}}{{\sqrt a - 1}}\\
b)A > 0\\
\Leftrightarrow \dfrac{{\sqrt a + 1}}{{\sqrt a - 1}} > 0\\
\Leftrightarrow \sqrt a - 1 > 0\\
\Leftrightarrow \sqrt a > 1\\
\Leftrightarrow a > 1\\
Vậy\,a > 1\\
c)\dfrac{1}{A} = \dfrac{{\sqrt a - 1}}{{\sqrt a + 1}} = \dfrac{{\sqrt a + 1 - 2}}{{\sqrt a + 1}}\\
= 1 - \dfrac{2}{{\sqrt a + 1}}\\
\sqrt a + 1 \ge 1\\
\Leftrightarrow \dfrac{2}{{\sqrt a + 1}} \le 2\\
\Leftrightarrow - \dfrac{2}{{\sqrt a + 1}} \ge - 2\\
\Leftrightarrow 1 - \dfrac{2}{{\sqrt a + 1}} \ge - 1\\
\Leftrightarrow \dfrac{1}{A} \ge - 1\\
\Leftrightarrow GTNN:\dfrac{1}{A} = - 1\,khi:a = 0
\end{array}$
